I was reading Wendl's notes on closed holomorphic curves, and could not seem to figure out the following assertion. I think I am missing some fact or the other about Riemann surfaces.
Let $\mathcal{M}_{g,m}$ be the space of pointed Riemann surfaces. To construct this, take the space of tuples $(\Sigma, j, \Theta)$ where $\Sigma$ is a topological closed surface of genus $g$, $j$ is a complex structure, and $\Theta$ is an ordered tuple of $m$ points in $\Sigma$. Then, quotient out by the equivalence relation $(\Sigma, j, \Theta) \sim (\Sigma', j', \Theta')$ if and only if there exists some biholomorphic diffeomorphism $\phi: \Sigma \to \Sigma'$ such that $\phi$ maps $\Theta$ to $\Theta'$ in an order-preserving manner.
We can define a related space as follows. Fix a surface $\Sigma$ and a tuple of $m$ points $\Phi$. Let $\text{Diff}_+(\Sigma, \Phi)$ be the set of all orientation-preserving diffeomorphisms on $\Sigma$ that restrict to the identity map on $\Phi$. Furthermore let $J(\Sigma)$ be the space of complex structures on $\Sigma$. Then one can take the quotient $J(\Sigma)/\text{Diff}_+(\Sigma, \Phi)$.
The claim is that $\mathcal{M}_{g,m}$ is homeomorphic to $J(\Sigma)/\text{Diff}_+(\Sigma, \Phi)$, likely by the map $[(\Sigma, j, \Theta)] \to [j]$ in one direction and $[j] \to [(\Sigma, j, \Phi)]$ in the opposite direction.
Right off the bat, to show that the forward map is injective, we must show that, for any complex structure $j'$ and points $\Theta$, the tuple $(\Sigma, j', \Theta)$ is equivalent to $(\Sigma, j', \Phi)$. This reduces to the following statement:
There exists a biholomorphic automorphism on $\Sigma$ that takes the set of points $\Theta$ to $\Phi$?
I am confused because this statement is certainly not true in general. For example, if we take $g = 0$ and $m = 4$, and $j$ the standard complex structure on $\Sigma = S^2$, the biholomorphic automorphisms are Mobius transformations. These are determined by the image of $\{0, 1, \infty\}$, so the action of the Mobius transformations is not transitive on quadruples of points.
Somewhat embarrassingly, this confusion was due to a failure on my part to absorb the necessary definitions.
There is always an orientation-preserving diffeomorphism from $\Sigma \setminus \Phi$ to $\Sigma \setminus \Theta$ for any set $\Theta$ of the same size, by virtue of the two surfaces being trivially homeomorphic.
This makes the fact that these two maps form a homeomorphism immediate. Any class in $\mathcal{M}_{g,m}$ admits a representative of the form $(\Sigma, j, \Phi)$, and in fact the composition of the two maps is the endomorphism on $\mathcal{M}_{g,m}$ that takes $[(\Sigma, j, \Phi)]$ to itself.
The other direction is clear by definition.